A Hooke's law spring is compressed 14.6 cm from equilibrium,and the potential energy stored is 76.0 J.What compression (as measured from equilibrium)would result in 100 J being stored in this case?
A)11.8 cm
B)16.7 cm
C)43.8 cm
D)33.5 cm
E)No answer is correct.

Question

Quizplus · Accepted Answer

The potential energy stored in a spring according to Hooke's law is given by $PE = \frac{1}{2}kx^2$, where $PE$ is the potential energy, $k$ is the spring constant, and $x$ is the compression or extension from equilibrium. Given that 76.0 J of energy is stored at a compression of 14.6 cm, we can find the spring constant $k$ and then use it to find the compression needed for 100 J.First, convert 14.6 cm to meters (since energy in Joules corresponds to meters in the formula): 14.6 cm = 0.146 m.Given $76.0 = \frac{1}{2}k(0.146)^2$, we solve for $k$.To find the compression for 100 J, set $100 = \frac{1}{2}kx^2$ and solve for $x$, using the same $k$.The calculations will show that the compression needed to store 100 J is greater than 14.6 cm but follows the relationship $x = \sqrt{\frac{2PE}{k}}$, leading to the correct answer being 16.7 cm, as it is the only option that is logically greater than 14.6 cm and fits the increase in energy stored.

A Hooke's Law Spring Is Compressed 14